Numbers With Same Consecutive Differences

class Solution {
    
    public int[] findNumbersWithConsecutiveDifferences(int len, int diff) {
        if (len == 1)
            return new int[] {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
    
        List<Integer> initialList = Arrays.asList(1, 2, 3, 4, 5, 6, 7, 8, 9);
        for(int i = 1; i < len; i++) {
            ArrayList<Integer> tmpList = new ArrayList<>();
            for(Integer number : initialList) {
                Integer lastDigit = number % 10;
    
                ArrayList<Integer> possibleDigits = new ArrayList<>();
                possibleDigits.add(lastDigit + diff);
                if (diff != 0)
                    possibleDigits.add(lastDigit - diff);
                for(Integer newDigit : possibleDigits) {
                    if (0 <= newDigit && newDigit < 10) {
                        Integer newNumber = number * 10 + newDigit;
                        tmpList.add(newNumber);
                    }
                }
            }
            initialList = tmpList;
        }
    
        return initialList.stream().mapToInt(j->j).toArray();
    }
}

This Java solution tackles the problem of finding numbers with specific consecutive digit differences. By defining a function findNumbersWithConsecutiveDifferences which accepts a length len for the number and a difference diff between two consecutive digits, the implementation explores all possible numbers that meet the criteria.

Here’s a breakdown of the approach:

• Begin with a list of single-digit numbers 1 through 9 since leading zeros are not allowed. If the len is specified as 1 (meaning a single-digit number), directly return all digits from 0 to 9.
• Iterate through each digit position, up to the length len. In each iteration, progressively construct numbers by adding the next possible valid digit.
• For each number derived from the previous iteration, ascertain the last digit and calculate potential next digits by adding and subtracting the given diff. Ensure these potential digits range from 0 to 9 to be valid for the next number part.
• Assemble these new valid numbers and proceed to the next iteration, progressively building the sequence as per the specified length.
• Convert the final list of valid numbers to an array before returning it, fulfilling the requirement for the output format.

This method ensures that all generated numbers are valid considering the required consecutive differences, harnessing the power of numbers' mathematical properties and list manipulation in a clear and structured manner. Given the respective constraints, the resulting implementation is efficient and can easily handle reasonable inputs of length and differences.

class Solution:
    def numberWithConsecutiveDiff(self, Len: int, Diff: int) -> List[int]:
        if Len == 1:
            return [i for i in range(10)]
    
        current_list = [num for num in range(1, 10)]
    
        for stage in range(Len-1):
            future_list = []
            for element in current_list:
                last_digit = element % 10
                possible_digits = set([last_digit + Diff, last_digit - Diff])
    
                for new_digit in possible_digits:
                    if 0 <= new_digit <= 9:
                        new_number = element * 10 + new_digit
                        future_list.append(new_number)
            current_list = future_list
    
        return current_list

This solution solves the problem of generating numbers that have the specified length Len, where each number's consecutive digits differ by a given difference Diff. The function numberWithConsecutiveDiff starts by handling the edge case where the length of the number is 1, returning all digits from 0 to 9 since they all satisfy the condition trivially.

For longer numbers (length greater than 1), the initial set of numbers (current_list) consists of the digits from 1 to 9 (omitting 0 as it cannot be a leading digit). The code then iterates over the desired length minus one (since the first digit is already added) to construct possible numbers:

• For each number in the current_list, extract the last digit.
• Determine the next possible digits by adding and subtracting Diff from the last digit.
• Filter these digits to ensure they are within the valid range (0 to 9).
• For each valid next digit, construct a new number by appending this digit to the current number (element) and multiply the current element by 10 to shift digits left before adding the new digit.
• Continue to build the list of valid numbers incrementally until the list is populated for the given length (Len).

The final list, current_list, contains all numbers of length Len where each pair of consecutive digits differs by the specified Diff. This solution employs list comprehensions for clear and concise collection of desired values and uses a loop with conditions to ensure the digits are within the bounds and check validity efficiently.